The corresponding equation in q is

The sign scheme for (3) is as shown

(3) q can not lie between
Subsequently:

a)      If a>1 then system (2) has no solutions, and therefore the original equation has no solution.

b)      If a=1, then system (2) has only unique solution i.e. x= -1 and the conditions of the original equation are not satisfied. Hence the original equation has no solution.

c)      If 0<a<1 then -1<-a<0 and therefore the interval [-a-2, -a] contains no less than four integers provided the inequality -a-2£ -4 holds true. Now solve the system.





Fig (16)
Thus , then the given equation has no less than four different integer solutions.
d)      If -1<a<0 then 0<-a<1 and therefore the interval [-a-2, -a] contains no less than four integer provided the inequality -a-2 £ -3 holds true. Now solve the system.

Thus , then the given equation has no less than four different integer solutions.

Q-2: Find all values of a for which the equation  has no less than four different integer solution.
Solution: The given equation can be written in the form,



The given equation is equivalent to the system,


The equation |A|+|B| = A-B

Holds true A 0 and B 0

Now consider following cases

Case 1: If a=0

Then system (1) gives equation have all xÎR as their solutions.


Fig (17)
Case 2: If a ¹0 then system (1) is equivalent to
 
Now,


For a<1, a 0

-a-2+a<0
For a=1
-a-2+a=0
For a>1
-a-2+a>0

Fig (18)
e)      If a= -1 then the interval [-1, 1] contains only three integer i.e. condition of the   problem are not satisfied.
f)        If a<-1 then -1<-a-2<0 and therefore the interval [-a-2, -a] contains no less than four integers. It is necessary that the inequality –a ³ 3 hold true thus for a £ -3 the given equation has no less than four integer solutions.
Combining all the results we get the set of required values for a,

Q-3: Solve for x

Solution:
Here,

• Quadratic


• Root


• Discreminant


• Polynomial


• Coefficient


• Common roots


• Lagrange’s Identity


• Cauchy Swartz Inequality


• Multiplicity